Question

17. As observed from the top of a 150 m high lighthouse from the sea-level, the angles of depression
of two ships are 30º and 45º. If one ship is exactly behind the other on the same side of the
lighthouse, find the distance between the two ships.​

Answer 1

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Answer 2

The distance between the two ships is 109.8 m.

Explanation:

According to the given description, we can make a diagram (Given in attachment), where AB =  lighthouse.

C and D are the position of two ships respectively.

Δ ABC and ΔABD are rigt triangles as the ligght house is standing vertical making 90° angle.

Consider triangle ABC , by trigonometry we have

$\tan 45^{\circ}=\dfrac{AB}{BC}$

$\Rightarrow\ 1=\dfrac{150}{BC}\Rightarrow BC = 150\ m$

Consider triangle ABD , we have

$\tan 30^{\circ}=\dfrac{AB}{BD}$

$\Rightarrow\ \dfrac{1}{\sqrt{3}}=\dfrac{150}{BD}\Rightarrow BD = 150\sqrt{3}\ m$

Now , CD = BD- BC

$=150\sqrt{3}-150=150(\sqrt{3}-1)\ m$

Put value of $\sqrt{3}=1.732$ , we get

CD = $150(1.732-1)= 109.8\ m$

Hence, the distance between the two ships is 109.8 m.

# Learn more :

From the top of a lighthouse 100m high the angles of depression of two ships on opposite sides of it are 48° and 36° respectively.Find the distance between the two ships

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